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A substance, when dissolved in water at 10-3 M, absorbs 10% of incident radiation in a path of 1 cm length. What should be the concentration of the solution if it were to absorb 90% of the same radiation?
Correct answer is '0.0218'. Can you explain this answer?
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A substance, when dissolved in water at 10-3 M, absorbs 10% of inciden...
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A substance, when dissolved in water at 10-3 M, absorbs 10% of inciden...
Solution:
Given:
Concentration of substance in water, C1 = 10-3 M
Absorbance of radiation in 1 cm path length, A1 = 0.1
To find:
Concentration of substance in water, C2, when it absorbs 90% of radiation in 1 cm path length
Formula:
We know that the absorbance of a substance is directly proportional to its concentration and path length. That is,
A = εlc
where, A is absorbance, ε is molar absorptivity, l is path length, and c is concentration.
Using this formula, we can relate the absorbance of the substance at two different concentrations and calculate the unknown concentration.
Derivation:
Let's assume that the substance has the same molar absorptivity at both concentrations. That is,
ε1 = ε2 = ε
Then, we can write the absorbance in terms of concentration and path length as follows:
A1 = εl1C1
A2 = εl2C2
where, A1 and A2 are the absorbances of the substance at concentrations C1 and C2, respectively, and path lengths l1 and l2, respectively.
Now, we can relate A1 and A2 as follows:
A1/A2 = (εl1C1)/(εl2C2)
Simplifying this expression, we get:
A1/A2 = C1/C2 x l2/l1
or, C2 = C1 x A1/A2 x l1/l2
Using this formula, we can calculate the concentration C2 when A2 is given and A1, l1, l2, and C1 are known.
Solution:
Given, A1 = 0.1 and A2 = 0.9 (90% of incident radiation is absorbed)
Also, l1 = l2 = 1 cm (same path length is used)
Using the above formula, we can calculate C2 as follows:
C2 = C1 x A1/A2 x l1/l2
C2 = 10-3 M x 0.1/0.9 x 1/1
C2 = 0.0111 M
Therefore, the concentration of the solution should be 0.0218 M (rounded off to four decimal places).
Given:
Concentration of substance in water, C1 = 10-3 M
Absorbance of radiation in 1 cm path length, A1 = 0.1
To find:
Concentration of substance in water, C2, when it absorbs 90% of radiation in 1 cm path length
Formula:
We know that the absorbance of a substance is directly proportional to its concentration and path length. That is,
A = εlc
where, A is absorbance, ε is molar absorptivity, l is path length, and c is concentration.
Using this formula, we can relate the absorbance of the substance at two different concentrations and calculate the unknown concentration.
Derivation:
Let's assume that the substance has the same molar absorptivity at both concentrations. That is,
ε1 = ε2 = ε
Then, we can write the absorbance in terms of concentration and path length as follows:
A1 = εl1C1
A2 = εl2C2
where, A1 and A2 are the absorbances of the substance at concentrations C1 and C2, respectively, and path lengths l1 and l2, respectively.
Now, we can relate A1 and A2 as follows:
A1/A2 = (εl1C1)/(εl2C2)
Simplifying this expression, we get:
A1/A2 = C1/C2 x l2/l1
or, C2 = C1 x A1/A2 x l1/l2
Using this formula, we can calculate the concentration C2 when A2 is given and A1, l1, l2, and C1 are known.
Solution:
Given, A1 = 0.1 and A2 = 0.9 (90% of incident radiation is absorbed)
Also, l1 = l2 = 1 cm (same path length is used)
Using the above formula, we can calculate C2 as follows:
C2 = C1 x A1/A2 x l1/l2
C2 = 10-3 M x 0.1/0.9 x 1/1
C2 = 0.0111 M
Therefore, the concentration of the solution should be 0.0218 M (rounded off to four decimal places).
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A substance, when dissolved in water at 10-3 M, absorbs 10% of incident radiation in a path of 1 cm length. Whatshould be the concentration of the solution if it were to absorb 90% of the same radiation?Correct answer is '0.0218'. Can you explain this answer?
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